From the previous problem, the height of each window is 1 m and the space betwee

Question

From the previous problem, the height of each window is 1 m and the space between windows is also 1 m. The tangerine was moving up with 12 m/s at the bottom of window #3. Calculate the average velocity of the tangerin while it is passing each window b Calculate time the tangerin takes to pass each window. Calculate the change in the velocity during each passing. Convert the running time of the second track runner from min to s. Convert the length of the first track from km to m. The average speed of both the runners is the same. Therefore, the difference of the lengths of tracks 1 and 2 is as follows. Substitute Rounding off to two significant figures, the difference of the lengths of the tracks 1 and 2 is 1.4 m If the difference between the lengths is not more than 1.4 m, then runner 1 is faster than runner 2. When the length of the track 1 is shorter than the length of the track 2 by more than 1.4 m, then runner 2 would definitely be faster.

Explanation / Answer

The initial velocity u = 12 m/s

A

Velocity of the tangerine after passing a window can be found using V2 = u2 -2gs

Let v1, v2, v3 be the velocities while crossing So

v12 = 122 – 2 x 9.8 x 1

v1= 11.15 m/s

v22 = 122 – 2 x 9.8 x 3

v2= 9.23 m/s

v32 = 122 – 2 x 9.8 x 5

v3= 6.78 m/s

B)

V= u – gt

u-v = gt

t = (u-v)/g

t1 = (u – v1)/g

= (12 - 11.15)/ 9.8

= 0.0867 s

t2 = (u – v2)/g

= (12 - 9.23)/9.8

= 0.2826 s

t3 = (u – v3)/g

= (12 – 6.78)/9.8

= 0.5326 s

C)

Change in velocity duting each passing are

U - v1 = 0.85 m/s

v1 -v2 = 1.92 m/s

v2 - v3 = 2.44 m/s

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